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Integral Calculus

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29 Oct 2007 21:45:26 IST

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das guptas problem

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if lim(x-0)f(x) exists where f(x) exists where f(x)=sin(3p-1)x/3x,x<0

tan(3p+1)x/2x,x>0 then p=

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Bipin Dubey

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Joined: 23 Jan 2007

Posts: 7942

30 Oct 2007 02:26:29 IST

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Since the limit exists LHL = RHL

LHL = _[x]

_[0-] f(x)

For x<0 f(x) = sin(3p-1)x/3x

Hence LHL = _[x]

_[0-] sin(3p-1)x/3x = (3p-1)/3

LHL = _[x]

_[0+] f(x)

For x>0 f (x) = tan(3p+1)x/2x

RHL = _[x]

_[0+] tan(3p+1)x/2x = (3p+1)/2

Since LHL = RHL

(3p-1)/3 = (3p+1)/2

p = -5/3

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